a) \(5x^2-7x-12=0\)
Ta có: \(a=5;b=-7;c=-12\)
\(\Rightarrow\Delta=b^2-4ac=\left(-7\right)^2-4.5.\left(-12\right)=289\)
Vậy phương trình có 2 nghiệm:
\(\left\{{}\begin{matrix}x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-\left(-7\right)-\sqrt{289}}{2.5}=-1\\x_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-\left(-7\right)+\sqrt{289}}{2.5}=2,4\end{matrix}\right.\)